January 2016

[R. Gerbicz]

The new Ibm puzzle is out: https://www.research.ibm.com/haifa/p...nuary2016.html
Note that the December 2015 puzzle is still running!

ps. Somewhat similar to the currently running azspcs contest (http://azspcs.net/Contest/SumsAndProducts1), Though I don't prefer these contests from Al. He is a world leading expert to post only(!) trivial (easy even for small kids) contests. See the totally boring stat:
Rank Score Contestant Last Improvement
1 25.00000 Tomas Rokicki Palo Alto, California, United States 19 Nov 2015 22:49
2 25.00000 Oleg Vlasii Ivano-Frankivsk, Ukraine 19 Nov 2015 22:54
3 25.00000 Kevin Modzelewski San Francisco, California, United States 21 Nov 2015 03:08
4 25.00000 Klaus Müller Koblenz, Germany 21 Nov 2015 10:01
5 25.00000 Nicholas Jimsheleishvili San Francisco, California, United States 26 Nov 2015 01:39
6 25.00000 Dmitry Kamenetsky Adelaide, Australia 26 Nov 2015 03:59
7 25.00000 Klaus & Lars Nagel München & Mainz, Germany 27 Nov 2015 03:54
8 25.00000 Roy van Rijn Maassluis, Netherlands 27 Nov 2015 07:34
9 25.00000 Moritz Franckenstein Minden, Nordrhein-Westfalen, Germany 28 Nov 2015 13:47
10 25.00000 Jarek Wroblewski Wroclaw, Poland 28 Nov 2015 19:45
11 25.00000 Martin Piotte Montreal, Quebec, Canada 30 Nov 2015 11:47
12 25.00000 Markus Sigg Freiburg, Germany 1 Dec 2015 15:54
13 25.00000 Johan Roos Lund, Sweden 3 Dec 2015 07:42
14 25.00000 Lucien Pech Zürich, Switzerland 9 Dec 2015 17:28
15 25.00000 Hans-Bernhard Meyer Au, Germany 22 Dec 2015 08:26
16 25.00000 Walter Trump Nuremberg, Germany 23 Dec 2015 16:17

When he recognised that his contest is totally trivial he posted another boring contest with the following stat:
Rank Score Contestant Last Improvement
1 30.0000 Tomas Rokicki Palo Alto, California, United States 2 Dec 2015 05:54
2 30.0000 Gil Dogon Jerusalem, Israel 2 Dec 2015 19:37
3 30.0000 Hanhong Xue Fuzhou, China 4 Dec 2015 12:25
4 30.0000 Jarek Wroblewski Wroclaw, Poland 5 Dec 2015 10:06
5 30.0000 Jeremy Sawicki Menlo Park, California, United States 6 Dec 2015 07:08
6 30.0000 Oleg Vlasii Ivano-Frankivsk, Ukraine 7 Dec 2015 00:38
7 30.0000 Martin Piotte Montreal, Quebec, Canada 8 Dec 2015 02:51
8 30.0000 Wes Sampson La Jolla, California, United States 8 Dec 2015 19:59
9 30.0000 Herbert Kociemba Darmstadt, Germany 12 Dec 2015 15:05
10 30.0000 Michael Hürter Saarbrücken, Germany 18 Dec 2015 19:26
11 30.0000 Lucien Pech Zürich, Switzerland 18 Dec 2015 22:38
12 30.0000 George W. Barnett Davis, California, United States 19 Dec 2015 09:03
13 30.0000 John Morris Simi Valley, California, United States 22 Dec 2015 05:17
14 30.0000 Helge Keller Karlsruhe, Germany 27 Dec 2015 08:09
15 30.0000 Peter Mobb Eckmannshausen, Germany 28 Dec 2015 13:03
16 30.0000 Yurii Sigolaev Saint Petersburg, Russia 28 Dec 2015 18:44

[science_man_88]

we the first thing to know is what multiple of 3 do we want to go up to because the largest spacing we can have is 3 57=3*19 so we need to cover all numbers 2 mod 3 up to 56 at minimum include 1 in the first set we know there are 7 primes less than 18 so those multiple can't all be in one set and the sets have to have multiples that are get covered. so say our minimum set to cover is 2,5,8,11,14,17,20,23,26,29,32,35,38,41,44,47,50,53,56 factoring them we get: 2,5,2^3,11,2*7,17,4*5,23,2*13,29,2^5,5*7,2*19,41,4*11,47,2*25,53,2*28 so our set naively is {2,5,11,7,17,4,23,13,29,19,41,47,25/10,53,28/14} and this contains too many primes to work easily/at all so the minimum may not work that simply . if we shift everything to 0 mod 3 and append 1 we get {1,3,2*3,3*3,4*3,5*3,6*3,7*3,8*3,9*3,10*3,11*3,12*3,13*3,14*3,15*3,16*3,17*3,18*3,19*3} -> unknown I see I could use {3,9,18} {6,9,15} to cover a third of them but that's using their example for N=19 as a cheat. edit: never mind made an error I'm an idiot multiplying both by three multiplied everything by nine not 3 so I can find 2, 6 sets that cover most. just concat what I did on the original example and you get 1->19 and 3*(1->19) and some more.

[KangJ]

I am not sure that using rational numbers is better than using integer numbers.

I guess rational numbers can be used to optimize the solution. (To get more asterisk - How many asterisks are possible?)

However, in the problem, there is no constraints on the types of number. So I think using rational numbers is fine.

[axn]

These are cogs in gear. They must be integers.

----

Brute forcing in a narrow search space gives two solutions for 57. Searching higher.

[R. Gerbicz]

There was a change in the problem's description: "Earn an asterisk for a significant increase of N."
,so my single * doesn't show that I have reached only the slightly better N=57.

[axn]

Originally Posted by R. Gerbicz

There was a change in the problem's description: "Earn an asterisk for a significant increase of N."
,so my single * doesn't show that I have reached only the slightly better N=57.

I did a brute force search with the max allowed element of 29. It found nothing better than the two 57s. That probably means this is the max possible (to hit 58, you need to either hit 57 or 58, since hitting 59 directly is out of the question).

I am also curios why others didn't get a *. There are no solutions that is exactly 56. So presumably they didn't get it because it was not original?!

[R. Gerbicz]

Originally Posted by axn

So presumably they didn't get it because it was not original?!

No. Think about what KangJ writes: "However, in the problem, there is no constraints on the types of number. So I think using rational numbers is fine."

[cuBerBruce]

I found a solution for N=57 using random brute force searching and where I could fix the lowest element(s) in each set, and the maximum value. (integer-only solution)

I am guessing axn's other solution is not simply swapping S_1 and S_2, but a truly distinct solution.

[petrw1]

They say:

We will post the names of those who submit a correct, original solution!

Now since there "seem to be" only 2 solutions for N=57 with integers that would propose there are only at most 3 original solutions:

The first 57
The second 57
Both 57's

However there are 5 names posted.

I surmise from that there there are either:
- more 57's
- some higher than 57
- some non-integer solutions (though if we go with the initial definition there can only be integer solutions ... ):

As part of IBM's efforts to make a better and greener planet, 
this month's challenge is to design an efficient gear system for bicycles. 
The front chainrings (S_1) and back cogset (S_2) ...

unless you can actually build a functioning bicycle with a fraction of a gear.

[petrw1]

Originally Posted by cuBerBruce

I am guessing axn's other solution is not simply swapping S_1 and S_2, but a truly distinct solution.

YES

[R. Gerbicz]

Originally Posted by petrw1

They say:

We will post the names of those who submit a correct, original solution!

Now since there "seem to be" only 2 solutions for N=57 with integers that would propose there are only at most 3 original solutions:

The first 57
The second 57
Both 57's

However there are 5 names posted.

Check out your dictionary for the word "original" or "original solution" (try https://translate.google.com). Or you can see some previous problems: https://www.research.ibm.com/haifa/p...s/May2015.html
here there was only 3 solutions, but much more solvers.

My own definition for an original solution: not other people's copy-pasted/stolen solution. And my wild guess is that for N=57 you will not get a star.

Originally Posted by petrw1

unless you can actually build a functioning bicycle with a fraction of a gear.

Maybe it was a little unfortunate wording in the puzzle.